We know that a 1-form $\omega$ on a manifold $M$ is exact if and only if $\int_{\gamma}\omega=0$ for any closed loop $\gamma$. How can I prove the following generalization: $\omega$ is an exact n-form on $S^n$ if and only if $\int_{S^n}\omega=0$? One direction follows clearly by Stokes, but I am not sure how to generalize the first fact to prove the remaining direction. Thanks!

1 Answer
1

You need to prove somehow that $H^n_{deRham}(S^n)\cong\mathbb{R}$ (and that the isomorphism is given by the integral over $S^n$). One possibility is induction and Mayer-Vietoris sequence. Here is another, somewhat more geometrical way. If $g:S^n\to S^n$ is a rotation and $\beta\in \Omega^n(S^n)$ then $g^*\beta-\beta$ is exact (since if $f_1$ is homotopic to $f_2$ then $f_1^*=f_2^*$ on cohomology). When we average over $SO(n)$, we can see that any $n$-form on $S^n$ is cohomologous to a $SO(n)$-invariant $n$-form. Up to multiple there is only one such form - the volume form $\omega$. Any $n$-form is thus of the form $d\alpha+c\omega$, and your claim follows.

(this argument shows that to find de Rham cohomology of a homogeneous space of a connected compact Lie group, we can restrict ourselves to the sub-complex of invariant forms. If the space is symmetric then all invariant forms are closed, i.e. the cohomology is equal to the space of invariant forms.)

How can you conclude from induction and the Mayer-Vietoris that the map that gives the isomorphism is actually integration over $S^n$?
–
ManuelJun 3 '11 at 22:39

Ah, ok. From Mayer Vietoris we get $H^n(S^n)=\mathbb{R}$. Since $S^n$ is orientable there is a non vanishing form $\omega$, so $[\omega]$ spans $H^n(S^n)$. In particular, we can take $\omega$ such that its integral over $S^n$ is 1, thus integration over $S^n$ gives us a surjective map $H^n(S^n) \rightarrow \mathbb{R}$, so by dimensional considerations this map must be injective.
–
ManuelJun 3 '11 at 23:19